Integrated Rate Laws — Use integrated rate laws for 0th, 1st, and 2nd order reactions - General Chemistry 2

SECTION 1

Historical Context & Motivation

The study of how fast chemical reactions proceed — chemical kinetics — has roots extending back to the mid-nineteenth century, when chemists first began to systematically measure the rates at which reactants are consumed and products are formed. While equilibrium thermodynamics could predict whether a reaction is spontaneous, it remained silent on the crucial question of how quickly that transformation would occur. This gap motivated a new branch of physical chemistry that sought to express concentration as a function of time, ultimately giving rise to the integrated rate laws that remain central to modern kinetics.

1850 • Wilhelmy's Sucrose Hydrolysis

Ludwig Wilhelmy published the first quantitative kinetics study, measuring the acid-catalyzed hydrolysis of sucrose with a polarimeter. He demonstrated that the rate of disappearance of sucrose was proportional to its instantaneous concentration — the earliest documented first-order rate law.

1864 • Guldberg & Waage's Law of Mass Action

Cato Guldberg and Peter Waage formalized the relationship between reaction rate and concentration, establishing the law of mass action. Their work provided the theoretical foundation for writing differential rate expressions that could then be integrated.

1884 • van 't Hoff's Études de Dynamique Chimique

Jacobus van 't Hoff systematically classified reactions by their reaction order and derived explicit integrated forms for zero-, first-, and second-order kinetics. His 1884 treatise is widely regarded as the founding text of modern chemical kinetics.

1889 • Arrhenius Equation

Svante Arrhenius introduced his famous temperature-dependence equation for rate constants, linking the rate constant k in integrated rate laws to activation energy and temperature, completing the kinetic framework.

The differential rate law tells us how fast a reaction proceeds at a given instant, but it does not directly reveal the concentration of a reactant at an arbitrary future time. Answering the practical question — "If I start with 0.50 M of reactant A, what concentration remains after 30 minutes?" — requires integration of the differential rate law. This mathematical operation transforms rate = f(concentration) into concentration = f(time), yielding equations that are indispensable for reactor design, pharmacokinetics, and environmental modeling.

SECTION 2

Core Principles & Definitions

Before deriving the integrated forms, it is essential to establish the conceptual vocabulary that underpins all rate-law analysis. A differential rate law expresses the instantaneous rate as a product of the rate constant and concentration terms raised to their respective orders: rate = k[A]ⁿ, where n is the reaction order with respect to A. The integrated rate law is obtained by separating variables and integrating from the initial concentration [A]₀ at time zero to [A] at time t. Each order yields a different functional form and, crucially, a different linear plot that serves as a diagnostic tool for determining the order experimentally.

Reaction Order (n)

The exponent to which the concentration of a reactant is raised in the rate law. It is determined experimentally, not from stoichiometry. Common values are 0, 1, and 2, though fractional and negative orders also occur.

Rate Constant (k)

The proportionality constant in the rate law whose units depend on the overall reaction order. For a zero-order reaction, k has units of M·s⁻¹; for first-order, s⁻¹; and for second-order, M⁻¹·s⁻¹.

Half-Life (t₁/₂)

The time required for the concentration of a reactant to fall to half its initial value. Its dependence on [A]₀ is a key distinguishing feature among orders: concentration-dependent for 0th and 2nd order, concentration-independent for 1st order.

Linear Plot Diagnostic

Plotting the appropriate function of [A] versus time yields a straight line whose slope is related to k. The choice of [A] vs. t, ln[A] vs. t, or 1/[A] vs. t reveals zero-, first-, or second-order behavior, respectively.

Pseudo-Order Simplification

When one reactant is present in large excess, its concentration remains essentially constant and is absorbed into the rate constant. A second-order reaction may then appear first-order (pseudo-first-order), simplifying analysis.

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KEY TAKEAWAY

Think of a differential rate law as the speedometer of a reaction — it tells you how fast you are going at a particular instant. The integrated rate law is more like the trip odometer combined with a GPS: it tells you where you are (concentration) at any point on the journey (time). Just as different vehicles (reaction orders) consume fuel at different mathematical rates, each order requires its own integration to obtain the concentration-time relationship.

SECTION 3

Visual Explanation — Concentration vs. Time Profiles

The most intuitive way to distinguish among the three common reaction orders is to examine their concentration-versus-time plots. The diagram below shows the decay of [A] over time for zero-order (linear decay), first-order (exponential decay), and second-order (hyperbolic decay) reactions, all starting from the same initial concentration and using rate constants chosen to make the differences visually clear. Notice how the zero-order reaction depletes the reactant in a finite, predictable time, whereas first- and second-order reactions approach zero asymptotically.

Comparison of [A] vs. time for the three common reaction orders starting from [A]₀ = 1.0 M. The violet line (zero-order) decreases linearly and reaches zero at a finite time. The cyan curve (first-order) decays exponentially, and the pink curve (second-order) decays more slowly at later times as 1/[A] increases linearly.

Several critical observations emerge from this plot. First, the zero-order curve is a straight line with a slope of −k, meaning that the reaction consumes the same absolute amount of reactant per unit time regardless of how much remains — a behavior characteristic of surface-catalyzed reactions that are saturated with substrate. Second, the first-order curve is concave upward and never truly reaches zero; its constant half-life means successive halving intervals are identical, which is the hallmark of radioactive decay and many unimolecular decompositions. Third, the second-order curve decays even more gradually at later times because the rate is proportional to [A]², and as [A] shrinks, the rate diminishes more steeply than in the first-order case, producing progressively longer half-lives.

SECTION 4

Mathematical Framework — Deriving the Integrated Rate Laws

Each integrated rate law is obtained by the same procedure: write the differential rate law −d[A]/dt = k[A]ⁿ, separate variables to isolate [A] on one side and dt on the other, then integrate both sides with the boundary condition [A] = [A]₀ at t = 0. The resulting expression for each integer order n = 0, 1, 2 is summarized below, together with its linear-plot form and half-life expression.

Zero-Order (n = 0)

ZERO-ORDER INTEGRATED RATE LAW

[A] = [A]₀ − kt

Derived from −d[A]/dt = k. Rearranging: d[A] = −k dt. Integrating: [A] − [A]₀ = −kt. Plot [A] vs. t for a straight line with slope = −k and intercept = [A]₀. Units of k: M·s⁻¹ (or M·time⁻¹).

ZERO-ORDER HALF-LIFE

t₁/₂ = [A]₀ / (2k)

The half-life depends on initial concentration: higher [A]₀ yields a longer half-life. Each successive half-life is shorter than the previous one.

First-Order (n = 1)

FIRST-ORDER INTEGRATED RATE LAW

ln[A] = ln[A]₀ − kt or equivalently [A] = [A]₀ e⁻ᵏᵗ

Derived from −d[A]/dt = k[A]. Separating: d[A]/[A] = −k dt. Integrating: ln[A] − ln[A]₀ = −kt. Plot ln[A] vs. t for a straight line with slope = −k. Units of k: s⁻¹ (or time⁻¹).

FIRST-ORDER HALF-LIFE

t₁/₂ = ln 2 / k ≈ 0.693 / k

The half-life is independent of [A]₀ — a unique and experimentally useful feature. This is why radioactive decay (a first-order process) has a fixed half-life regardless of the amount of material.

Second-Order (n = 2)

SECOND-ORDER INTEGRATED RATE LAW

1/[A] = 1/[A]₀ + kt

Derived from −d[A]/dt = k[A]². Separating: d[A]/[A]² = −k dt. Integrating: −1/[A] + 1/[A]₀ = −kt, which rearranges to the form above. Plot 1/[A] vs. t for a straight line with slope = +k. Units of k: M⁻¹·s⁻¹.

SECOND-ORDER HALF-LIFE

t₁/₂ = 1 / (k[A]₀)

The half-life is inversely proportional to [A]₀: as the reactant is consumed and [A] decreases, each successive half-life becomes longer. This contrasts sharply with zero-order (shorter successive half-lives) and first-order (constant half-life).

SECTION 5

Diagnostic Linear Plots — Determining Reaction Order

In practice, the reaction order is often unknown and must be determined experimentally. The standard approach is to collect concentration-time data and then construct three candidate plots: [A] vs. t, ln[A] vs. t, and 1/[A] vs. t. The plot that yields a straight line with the best R² value identifies the reaction order. The slope of that line gives the rate constant (with the appropriate sign convention), and the intercept gives the initial condition.

Top row: the three diagnostic linear plots. Only one will yield a straight line for a given data set, revealing the reaction order. Bottom: summary table of the y-axis variable, slope, intercept, and units of k for each order.

Property	Zero-Order	First-Order	Second-Order
Integrated Law	[A] = [A]₀ − kt	ln[A] = ln[A]₀ − kt	1/[A] = 1/[A]₀ + kt
Linear Plot	[A] vs. t	ln[A] vs. t	1/[A] vs. t
Slope	−k	−k	+k
Half-Life	[A]₀ / 2k	0.693 / k	1 / (k[A]₀)
t₁/₂ Trend	Decreases as [A] drops	Constant	Increases as [A] drops
Units of k	M·s⁻¹	s⁻¹	M⁻¹·s⁻¹

Comprehensive comparison of integrated rate law properties across reaction orders

SECTION 6

Worked Example — Determining Order and Rate Constant from Data

Consider the decomposition of nitrogen dioxide: 2 NO₂(g) → 2 NO(g) + O₂(g). A kinetics experiment at 300 °C yields the following data for [NO₂] over time:

Time (s)	[NO₂] (M)	ln[NO₂]	1/[NO₂] (M⁻¹)
0	0.0100	−4.605	100.0
50	0.00787	−4.845	127.1
100	0.00649	−5.038	154.1
200	0.00481	−5.337	207.9
300	0.00380	−5.573	263.2

Experimental kinetics data for NO₂ decomposition at 300 °C

Determining the Order and Rate Constant for NO₂ Decomposition

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Step 1 — Compute Diagnostic Quantities

For each data point, calculate ln[NO₂] and 1/[NO₂]. These have already been tabulated above. We will now assess which of the three diagnostic plots ([NO₂] vs. t, ln[NO₂] vs. t, 1/[NO₂] vs. t) yields the best straight line.
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Step 2 — Test for Zero-Order

Plotting [NO₂] vs. t: the points (0, 0.0100), (50, 0.00787), (100, 0.00649), (200, 0.00481), (300, 0.00380) show a curve, not a straight line. The intervals between successive concentration values are not equal (Δ[A] changes from 0.00213 to 0.00138 to 0.00168 to 0.00101). This rules out zero-order kinetics.

Not zero-order — [NO₂] vs. t is curved.
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Step 3 — Test for First-Order

Plotting ln[NO₂] vs. t: the points are (0, −4.605), (50, −4.845), (100, −5.038), (200, −5.337), (300, −5.573). Checking the increments: the change from 0 to 50 s is −0.240, from 50 to 100 s is −0.193, from 100 to 200 s (100 s interval) is −0.299, and from 200 to 300 s is −0.236. These are not equally spaced per unit time, so ln[NO₂] vs. t is not perfectly linear.

Not first-order — ln[NO₂] vs. t shows curvature.
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Step 4 — Test for Second-Order

Plotting 1/[NO₂] vs. t: the points are (0, 100.0), (50, 127.1), (100, 154.1), (200, 207.9), (300, 263.2). The change per second is approximately 0.54 M⁻¹·s⁻¹ for each interval: (127.1 − 100.0)/50 = 0.542, (154.1 − 127.1)/50 = 0.540, (207.9 − 154.1)/100 = 0.538, (263.2 − 207.9)/100 = 0.553. These are remarkably consistent, confirming linearity.

Second-order confirmed — 1/[NO₂] vs. t is linear.
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Step 5 — Extract the Rate Constant

From the second-order integrated rate law, 1/[A] = 1/[A]₀ + kt, the slope of the 1/[NO₂] vs. t line equals k. Using the endpoints: k = (263.2 − 100.0) M⁻¹ / (300 − 0) s = 163.2 / 300 = 0.544 M⁻¹·s⁻¹.

k = 0.544 M⁻¹·s⁻¹
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Step 6 — Calculate the Half-Life

For a second-order reaction, t₁/₂ = 1/(k[A]₀) = 1/(0.544 M⁻¹·s⁻¹ × 0.0100 M) = 1/0.00544 s = 184 s. Note that this half-life applies only at the beginning when [NO₂] = 0.0100 M; as the reaction progresses and [NO₂] drops, the half-life will increase.

t₁/₂ = 184 s (initial half-life)

SECTION 7

Strengths, Limitations & Common Pitfalls

Integrated rate laws are powerful analytical tools, but they come with assumptions and limitations that must be acknowledged to avoid erroneous conclusions. The following comparison highlights the practical strengths and constraints of each form, along with common mistakes students and researchers encounter when applying these equations.

Aspect	Strengths	Limitations / Pitfalls
Zero-Order	Simple linear algebra; easy to extract k. Useful for enzyme-saturated (Michaelis–Menten limit) and surface-catalyzed reactions.	Predicts negative concentrations beyond t = [A]₀/k; physically meaningless past that point. Only valid while the rate-limiting step is saturated.
First-Order	Constant half-life simplifies pharmacokinetic dosing. ln[A] plot is highly sensitive for order determination. Handles pseudo-first-order elegantly.	Requires careful measurement of [A] over several half-lives. Students often forget the natural log and try to plot log₁₀[A] instead, altering the slope by a factor of 2.303.
Second-Order	Captures bimolecular elementary reactions. 1/[A] plot has a positive slope (unique), making it easy to identify. Increasing half-life is a useful diagnostic.	The form 1/[A] = 1/[A]₀ + kt applies only when the reaction is second-order in a single reactant or when both reactant concentrations are equal. For A + B → products with [A]₀ ≠ [B]₀, a more complex integration is needed.
General	All three forms allow extraction of k from a single kinetics run. They connect directly to mechanism and half-life, bridging theory and experiment.	Assumes constant temperature (k is temperature-dependent). Assumes no significant reverse reaction. Side reactions or impurities can distort linearity of all diagnostic plots.

Strengths and limitations of integrated rate laws by reaction order

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COMMON MISTAKE ALERT

The stoichiometric coefficient of the reactant does not determine the reaction order — reaction order is an experimentally determined quantity. A reaction 2A → products could be zero-, first-, or second-order depending on the mechanism. Also, remember that the integrated rate law for a second-order reaction involving two different reactants at unequal initial concentrations requires a partial-fraction integration that yields a logarithmic expression, not the simple 1/[A] form. Always verify what 'second-order' means in context: second-order in one reactant, or first-order in each of two reactants.

SECTION 8

Connection to Advanced Theory — Beyond Integer Orders

The integrated rate laws for 0th, 1st, and 2nd order reactions form the foundation upon which more sophisticated kinetic analyses are built. In advanced physical chemistry and chemical engineering, one frequently encounters situations that extend well beyond these elementary cases, including fractional reaction orders, consecutive and parallel reactions, and temperature-dependent kinetics governed by the Arrhenius equation. The table below summarizes how the elementary integrated rate laws connect to these more advanced topics.

Elementary Concept	Advanced Extension	Key Difference / Addition
Integer order n = 0, 1, 2	Fractional and negative orders	Arise from complex multi-step mechanisms (e.g., chain reactions, enzyme kinetics). The general integrated form uses 1/(n−1) · [A]¹⁻ⁿ for n ≠ 1.
Single-reactant rate law	Two-reactant second-order (A + B → P)	Requires partial-fraction decomposition when [A]₀ ≠ [B]₀, yielding ln([B][A]₀ / [A][B]₀) = ([B]₀ − [A]₀)kt.
Fixed rate constant k	Temperature-dependent k via Arrhenius	k = A·exp(−Eₐ/RT) links the rate constant to activation energy and temperature. Combining with integrated rate laws enables non-isothermal reactor design.
Isolated reaction	Consecutive reactions (A → B → C)	Requires coupled differential equations. The concentration of intermediate B passes through a maximum described by the Bateman equations.
Irreversible reaction	Reversible reaction kinetics	The integrated form must include both forward and reverse rate constants: the system approaches equilibrium rather than going to completion.

How elementary integrated rate laws connect to advanced kinetic theory

In courses on reactor design (chemical engineering) and pharmacokinetics (pharmaceutical sciences), the integrated rate laws serve as building blocks for modeling continuous-flow and batch reactors. For instance, the design equation for a batch reactor is essentially the integrated rate law rearranged to solve for the time needed to achieve a desired conversion. Similarly, the first-order integrated rate law underpins the one-compartment pharmacokinetic model used to predict drug plasma concentrations after dosing. Mastering the derivations and diagnostic plots covered in this lesson provides the essential foundation for all of these advanced applications.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

A student measures the half-life of a reaction at two different initial concentrations and finds that the half-life doubles when the initial concentration is halved. What is the reaction order, and how do you know?

Show answer

The reaction is second-order. For a second-order reaction, t₁/₂ = 1/(k[A]₀), so the half-life is inversely proportional to the initial concentration. Halving [A]₀ doubles 1/[A]₀, and therefore doubles the half-life. This is the unique signature of second-order kinetics. By contrast, a zero-order half-life ([A]₀/2k) would halve when [A]₀ is halved, and a first-order half-life (0.693/k) would remain unchanged regardless of [A]₀.

Question 2: BASIC CALCULATION

The decomposition of hydrogen peroxide (H₂O₂) in aqueous solution is first-order with a rate constant k = 1.06 × 10⁻³ s⁻¹. If [H₂O₂]₀ = 0.200 M, what is the concentration of H₂O₂ after 600 s?

Show answer

Using the first-order integrated rate law: ln[A] = ln[A]₀ − kt. Substituting: ln[H₂O₂] = ln(0.200) − (1.06 × 10⁻³)(600) = −1.6094 − 0.636 = −2.2454. Exponentiating: [H₂O₂] = e⁻²·²⁴⁵⁴ = 0.106 M. Approximately 47% of the original H₂O₂ has decomposed in 600 seconds. We can verify this is reasonable by computing the half-life: t₁/₂ = 0.693 / (1.06 × 10⁻³) = 654 s, so 600 s is slightly less than one half-life, and indeed more than half the original amount remains.

Question 3: INTERMEDIATE

A certain reaction A → products has a rate constant k = 0.025 M·min⁻¹ and follows zero-order kinetics. Starting from [A]₀ = 0.500 M, (a) how long does it take for [A] to reach 0.100 M, and (b) what is the concentration after 25 minutes?

Show answer

(a) Using [A] = [A]₀ − kt, we solve for t: t = ([A]₀ − [A])/k = (0.500 − 0.100)/0.025 = 0.400/0.025 = 16.0 minutes. (b) After 25 minutes: [A] = 0.500 − (0.025)(25) = 0.500 − 0.625 = −0.125 M. A negative concentration is physically impossible, which means the reactant has been completely consumed before 25 minutes. The reactant reaches [A] = 0 at t = [A]₀/k = 0.500/0.025 = 20 minutes, so [A] = 0 M after 25 min. This illustrates a critical limitation of the zero-order integrated rate law: it must not be extrapolated beyond the time at which the reactant is fully depleted.

Question 4: APPLIED

A pharmaceutical company needs to determine the shelf life of a drug that degrades via first-order kinetics with k = 4.50 × 10⁻⁶ s⁻¹ at 25 °C. The drug is considered ineffective when its concentration drops below 90% of its original formulation. How many days is the shelf life?

Show answer

The drug is effective as long as [A] ≥ 0.90[A]₀. Using ln[A] = ln[A]₀ − kt with [A]/[A]₀ = 0.90: ln(0.90) = −kt, so t = −ln(0.90)/k = −(−0.10536)/(4.50 × 10⁻⁶) = 0.10536/(4.50 × 10⁻⁶) = 23,413 s. Converting to days: 23,413 s × (1 min/60 s) × (1 hr/60 min) × (1 day/24 hr) = 0.271 days ≈ 6.5 hours. This extremely short shelf life indicates the drug requires cold storage or a stabilizing formulation. In pharmaceutical practice, the Arrhenius equation would be used to determine how much lower the temperature must be to extend the shelf life to an acceptable duration, which is why integrated rate laws and temperature dependence are intimately connected.

Question 5: CRITICAL THINKING

A kinetics experiment yields the following data: at t = 0, [A] = 0.80 M; at t = 100 s, [A] = 0.40 M; at t = 300 s, [A] = 0.20 M. A student claims the reaction is first-order because the concentration halves. Critically evaluate this claim by testing the data against all three integrated rate laws and determining the true order.

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The student notices that [A] goes from 0.80 to 0.40 M (first halving) and from 0.40 to 0.20 M (second halving), but the time intervals for these halvings are different: 100 s for the first and 200 s for the second. For a true first-order reaction, the half-life is constant, so both halvings should take the same time. The increasing half-life (100 s, then 200 s — a doubling) is the hallmark of a second-order reaction, where t₁/₂ = 1/(k[A]). As [A] drops from 0.80 to 0.40 M (halved), t₁/₂ doubles from 100 to 200 s. We can verify: for second-order, k = 1/(t₁/₂ × [A]₀) = 1/(100 × 0.80) = 0.0125 M⁻¹s⁻¹. Check: 1/[A] at t = 0 is 1.25, at t = 100 is 2.50, at t = 300 is 5.00. Plotting: slope = (5.00 − 1.25)/300 = 0.0125 M⁻¹s⁻¹, perfectly linear. The reaction is definitively second-order, and this problem illustrates why simply observing 'halving' is insufficient — one must verify that the half-life is constant before concluding first-order kinetics.

SUMMARY

Lesson Summary

Summary

Integrated rate laws transform the instantaneous differential rate law into a concentration-versus-time expression by separating variables and integrating. For a zero-order reaction, [A] = [A]₀ − kt (linear decay, units of k: M·s⁻¹); for a first-order reaction, ln[A] = ln[A]₀ − kt (exponential decay, units of k: s⁻¹); and for a second-order reaction, 1/[A] = 1/[A]₀ + kt (hyperbolic decay, units of k: M⁻¹·s⁻¹). Each form corresponds to a unique diagnostic linear plot — [A] vs. t, ln[A] vs. t, or 1/[A] vs. t — whose linearity identifies the reaction order and whose slope yields the rate constant.

The half-life behavior provides an additional diagnostic: zero-order half-lives decrease as the reaction progresses, first-order half-lives remain constant, and second-order half-lives increase. These relationships follow directly from the mathematical dependence (or independence) of t₁/₂ on [A]₀. By mastering these three integrated forms, their derivations, and their graphical signatures, you possess the core analytical toolkit for experimental kinetics, with direct applications to reactor design, pharmacokinetics, environmental chemistry, and the Arrhenius analysis that connects rate constants to temperature and activation energy.